What is the real positive $\upsilon$ that satisfies $\upsilon^\upsilon=\upsilon+1$? I think the Lambert-W function might be relevant here, but I have no idea how to use it.
I just really like the letter upsilon. It doesn't get enough love.
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1Fyi, Mathematica 9 has no built-in methods to solve it symbolically. – alancalvitti Dec 21 '12 at 19:34
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2Newtons method? – picakhu Dec 21 '12 at 20:15
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2I think only numerical solution. – JACKY88 Dec 21 '12 at 20:23
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2No, LambertW says nothing about this. I doubt very much that there is a closed-form solution. – Robert Israel Dec 21 '12 at 20:39
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5$x^x = x + 2$ is easier. – Will Jagy Dec 21 '12 at 20:54
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1@Will Jagy: $x^x=x+1$ is also not difficult. Try $x=0$... – Fabian Dec 21 '12 at 21:46
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yeah, but I meant the positive one. – B H Dec 21 '12 at 23:42
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1That's only if you define $0^0$ to be $1$. Sometimes it is defined to be indeterminate. – toypajme Dec 21 '12 at 23:42
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1In this case, since $\lim_{x \to 0} x^x = 1$, I think we can safely define $0^0 = 1$ for this one thing. – Joe Z. Jan 05 '13 at 03:01
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It is possible to write this in terms of Lambert W, but it doesn't help in solving the problem at all. $ \ln \upsilon = \mathrm{W}(\ln(\upsilon + 1)) $ – George V. Williams Jan 06 '13 at 00:15
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@hombre How do you mean "closed form" solution? – vesszabo Jan 06 '13 at 13:45
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Heh, good question, vesszabo. This is a bit vague, but I was hoping for a solution in terms of functions that we already know. For example, the error function, Lambert-W, all of that good stuff. I think it might have to do with Galois Groups or something, whether it can be written in terms of functions we already know, or it has to be done in terms of something else. I'm not sure. – B H Jan 06 '13 at 15:56
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1The inverse symbolic calculator gives nothing useful. – mrf Jan 07 '13 at 09:00
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4See http://oeis.org/A124930 . According to this web page, this number is transcendental. Most likely, there is no closed form expression for it in terms of elementary functions. – Yury Jan 11 '13 at 02:24
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@BH Is it okay if one finds an expression for complex solutions? – Тyma Gaidash May 09 '23 at 20:59
2 Answers
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I used a very simple method, "iterative fixed point method", look that $x^{x}=x+1$ is equivalent to $x^{x}-x-1=0$ and $-2x=x^{x}-3x-1$ and $x=\frac{x^{x}-3x-1}{-2}$. Define $g(x)=\frac{x^{x}-3x-1}{-2}$ so the answer of equation $x^{x}=x+1$ is the fixed point of the function $g(x)$. For finding it we note that because $g'(x)=\frac{(1+ln(x))x^{x}-3}{-2}$ is decreasing in interval $[1.65,1.9]$ and $g'(1.65)=-0.2124....$ and $g'(1.9)=-0.667....$ so $g'(x)$ is negative in interval $[1.65,1.9]$ and $g(x)$ is decreasing in this interval and as $g(1.65)=1.83....$ and $g(1.9)=1.65....$ we have $$\forall x\in [1.65,1.9] \; :\;g(x)\in [1.65,1.9]$$ and also we have seen that $$\forall x\in [1.65,1.9] \; : \; |g'(x)|\leq 0.67<1$$ so by a theorem in Numerical Analysis for iterative fixed point method the sequence $\{g(x_{n})\}_{n=1}^{\infty}$ is convergence to requested point, if we choose $x_{1}\in [1.65,1.9]$. But if you want something else like only using "Lambert W function" and such things, please say me.
AmirHosein Sadeghimanesh
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@hombre, Perhaps You don't want numerical solutions and looking to this number as a limit of a sequence, from above discussions it is brightly that you want to earn this number as some form of some well-known functions. – AmirHosein Sadeghimanesh Jan 07 '13 at 08:21
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1Yeah, you're right. Thank you; I appreciate your answer, but I was hoping for a solution that wasn't a recursive sequence or an infinite sum or anything like that. – B H Jan 07 '13 at 22:41
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If you plan to use newton's method, it helps to simplify first: Take $\log$ on both sides to get $$ \upsilon \log(\upsilon) = \log(\upsilon+1)$$
Let $$f(\upsilon) = \upsilon \log(\upsilon) - \log(\upsilon+1)$$ Then $$ f'(\upsilon) = \log(\upsilon) - \frac{\upsilon}{\upsilon+1}$$ The Newton's method gives
$$\upsilon_{n+1} = \upsilon_n - \frac{f(\upsilon)}{f'(\upsilon)}$$
Starting with $$\upsilon_0 =1$$ one gets the solution to machine precision in 5 steps as $$\upsilon = 1.77677504009705$$
user44197
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